Comprehension
A hydrogen atom consists of an electron revolving in a circular orbit of radius r with certain velocity v around a proton located at the nucleus of the atom. The electrostatic force of attraction between the revolving electron and the proton provides the requisite centripetal force to keep it in the orbit. According to Bohr’s model, an electron can revolve only in certain stable orbits. The angular momentum of the electron in these orbits is some integral multiple of \(\frac{h}{2π}\), where h is the Planck’s constant.
Question: 1
The expression for the speed of electron \( v \) in terms of radius of the orbit (\( r \)) and physical constant \[ K = \frac{1}{4 \pi \varepsilon_0} \] is:
The expression for the speed of electron \( v \) in terms of radius of the orbit (\( r \)) and physical constant \[ K = \frac{1}{4 \pi \varepsilon_0} \] is:
Show Hint
Always equate centripetal force and electrostatic force in Bohr model derivations for electron motion in hydrogen-like atoms.
Updated On: Feb 9, 2026
- \( \frac{Ke^2}{mr} \)
- \( \frac{Ke^2}{mr^2} \)
- \( \sqrt{\frac{Ke^2}{mr}} \)
- \( \sqrt{\frac{Ke^2}{mr^2}} \)
Show Solution
The Correct Option is C
Solution and Explanation
- The Coulomb attractive force between the nucleus and the electron supplies the centripetal force necessary for the electron to maintain a circular orbit.
\[\frac{mv^2}{r} = \frac{Ke^2}{r^2}\]
- The electron's orbital velocity \( v \) is determined by solving the equation for \( v \):
\[mv^2 = \frac{Ke^2}{r} \Rightarrow v^2 = \frac{Ke^2}{mr}\Rightarrow v = \sqrt{\frac{Ke^2}{mr}}\]
\[\frac{mv^2}{r} = \frac{Ke^2}{r^2}\]
- The electron's orbital velocity \( v \) is determined by solving the equation for \( v \):
\[mv^2 = \frac{Ke^2}{r} \Rightarrow v^2 = \frac{Ke^2}{mr}\Rightarrow v = \sqrt{\frac{Ke^2}{mr}}\]
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Question: 2
The total energy of the atom in terms of \( r \) and physical constant \( K \) is:
The total energy of the atom in terms of \( r \) and physical constant \( K \) is:
Show Hint
Total energy in Bohr’s model is always negative, indicating a bound system. It's half the potential energy in magnitude but with opposite sign.
Updated On: Feb 9, 2026
- \( \frac{Ke^2}{r} \)
- \( -\frac{Ke^2}{2r} \)
- \( \frac{Ke^2}{2r} \)
- \( \frac{3}{2} \cdot \frac{Ke^2}{r} \)
Show Solution
The Correct Option is B
Solution and Explanation
- The total energy \( E \) of an electron in a hydrogen atom is the sum of its kinetic energy (K.E.) and potential energy (P.E.).
- Kinetic energy is given by:
\[ \text{K.E.} = \frac{1}{2}mv^2 \]
- Using the centripetal force equation, we previously found:
\[ v^2 = \frac{Ke^2}{mr} \]
Therefore, K.E. simplifies to:
\[ \text{K.E.} = \frac{1}{2} m \cdot \frac{Ke^2}{mr} = \frac{Ke^2}{2r} \]
- The potential energy between two opposite charges is:
\[ \text{P.E.} = -\frac{Ke^2}{r} \]
- The total energy is the sum of K.E. and P.E.:
\[ E = \text{K.E.} + \text{P.E.} = \frac{Ke^2}{2r} - \frac{Ke^2}{r} = -\frac{Ke^2}{2r} \]
- Kinetic energy is given by:
\[ \text{K.E.} = \frac{1}{2}mv^2 \]
- Using the centripetal force equation, we previously found:
\[ v^2 = \frac{Ke^2}{mr} \]
Therefore, K.E. simplifies to:
\[ \text{K.E.} = \frac{1}{2} m \cdot \frac{Ke^2}{mr} = \frac{Ke^2}{2r} \]
- The potential energy between two opposite charges is:
\[ \text{P.E.} = -\frac{Ke^2}{r} \]
- The total energy is the sum of K.E. and P.E.:
\[ E = \text{K.E.} + \text{P.E.} = \frac{Ke^2}{2r} - \frac{Ke^2}{r} = -\frac{Ke^2}{2r} \]
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Question: 3
A photon of wavelength 500 nm is emitted when an electron makes a transition from one state to the other state in an atom. The change in the total energy of the electron and change in its kinetic energy in eV as per Bohr’s model, respectively will be:
A photon of wavelength 500 nm is emitted when an electron makes a transition from one state to the other state in an atom. The change in the total energy of the electron and change in its kinetic energy in eV as per Bohr’s model, respectively will be:
Show Hint
Photon emission leads to decrease in total energy. The kinetic energy of the electron increases by the same amount due to the negative potential energy change.
Updated On: Feb 9, 2026
- \( 2.48, -2.48 \)
- \( -1.24, 1.24 \)
- \( -2.48, 2.48 \)
- \( 1.24, -1.24 \)
Show Solution
The Correct Option is C
Solution and Explanation
- The energy of an emitted photon is calculated using \( E = \frac{hc}{\lambda} \).
- Given constants are \( h = 6.63 \times 10^{-34} \, \text{Js} \), \( c = 3 \times 10^8 \, \text{m/s} \), and \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \).
\[ E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}} = 3.978 \times 10^{-19} \, \text{J} \]
- Converting this energy to electron volts: \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \).
\[ E = \frac{3.978 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.48 \, \text{eV} \]
- In Bohr's model, the change in total energy is \( \Delta E = -2.48 \, \text{eV} \) due to energy release.
- The change in kinetic energy is \( \Delta KE = -\frac{1}{2} \Delta PE = +2.48 \, \text{eV} \), as kinetic energy is \( -E \).
- Given constants are \( h = 6.63 \times 10^{-34} \, \text{Js} \), \( c = 3 \times 10^8 \, \text{m/s} \), and \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \).
\[ E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}} = 3.978 \times 10^{-19} \, \text{J} \]
- Converting this energy to electron volts: \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \).
\[ E = \frac{3.978 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.48 \, \text{eV} \]
- In Bohr's model, the change in total energy is \( \Delta E = -2.48 \, \text{eV} \) due to energy release.
- The change in kinetic energy is \( \Delta KE = -\frac{1}{2} \Delta PE = +2.48 \, \text{eV} \), as kinetic energy is \( -E \).
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Question: 4
In Bohr’s model of hydrogen atom, the frequency of revolution of electron in its \( n^\text{th} \) orbit is proportional to:
In Bohr’s model of hydrogen atom, the frequency of revolution of electron in its \( n^\text{th} \) orbit is proportional to:
Show Hint
In Bohr’s model, remember \( v \propto \frac{1}{n} \) and \( r \propto n^2 \). Use \( f = \frac{v}{2\pi r} \) to derive frequency dependence on \( n \).
Updated On: Feb 9, 2026
- \( n \)
- \( \frac{1}{n} \)
- \( \frac{1}{n^2} \)
- \( \frac{1}{n^3} \)
Show Solution
The Correct Option is D
Solution and Explanation
- The frequency of revolution \( f \) in Bohr's model is defined as:
\[
f = \frac{v}{2\pi r}
\]
- Based on Bohr's findings for velocity and radius: \[ v \propto \frac{1}{n}, \quad r \propto n^2 \] Therefore, \[ f \propto \frac{1/n}{n^2} = \frac{1}{n^3} \]
- Based on Bohr's findings for velocity and radius: \[ v \propto \frac{1}{n}, \quad r \propto n^2 \] Therefore, \[ f \propto \frac{1/n}{n^2} = \frac{1}{n^3} \]
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Question: 5
An electron makes a transition from \(-3.4\ \text{eV}\) state to the ground state in hydrogen atom. Its radius of orbit changes by: (radius of orbit of electron in ground state = \(0.53\ \text{\AA}\))
An electron makes a transition from \(-3.4\ \text{eV}\) state to the ground state in hydrogen atom. Its radius of orbit changes by: (radius of orbit of electron in ground state = \(0.53\ \text{\AA}\))
Show Hint
Use energy values to find quantum number \(n\), then apply \(r_n = n^2 \cdot a_0\) to calculate orbit radii in hydrogen-like atoms.
Updated On: Feb 9, 2026
- \(0.53\ \text{\AA}\)
- \(1.06\ \text{\AA}\)
- \(1.59\ \text{\AA}\)
- \(2.12\ \text{\AA}\)
Show Solution
The Correct Option is C
Solution and Explanation
- The energy of an electron in the \(n^{\text{th}}\) orbit of hydrogen is calculated as:
\[
E_n = -13.6 \frac{1}{n^2} \text{ eV}
\]
- For a given energy level \(E = -3.4\ \text{eV}\), the corresponding quantum number \(n\) is determined by: \[ -3.4 = -13.6 \cdot \frac{1}{n^2} \Rightarrow \frac{1}{n^2} = \frac{3.4}{13.6} = \frac{1}{4} \Rightarrow n = 2 \]
- The radius of an orbit in Bohr's model is given by: \[ r_n = n^2 \cdot a_0, \text{ where } a_0 = 0.53\ \text{\AA} \]
This results in a radius for the \(n=2\) orbit: \[ r_2 = 2^2 \cdot 0.53 = 4 \cdot 0.53 = 2.12\ \text{\AA} \]
- Comparing this to the radius of the ground state (\(r_1 = 0.53\ \text{\AA}\)), the change in radius is: \[ 2.12 - 0.53 = 1.59\ \text{\AA} \]
- For a given energy level \(E = -3.4\ \text{eV}\), the corresponding quantum number \(n\) is determined by: \[ -3.4 = -13.6 \cdot \frac{1}{n^2} \Rightarrow \frac{1}{n^2} = \frac{3.4}{13.6} = \frac{1}{4} \Rightarrow n = 2 \]
- The radius of an orbit in Bohr's model is given by: \[ r_n = n^2 \cdot a_0, \text{ where } a_0 = 0.53\ \text{\AA} \]
This results in a radius for the \(n=2\) orbit: \[ r_2 = 2^2 \cdot 0.53 = 4 \cdot 0.53 = 2.12\ \text{\AA} \]
- Comparing this to the radius of the ground state (\(r_1 = 0.53\ \text{\AA}\)), the change in radius is: \[ 2.12 - 0.53 = 1.59\ \text{\AA} \]
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